Pye and Associates recently conducted a survey to determine the percent of adult county residents who believe that students in public high schools should wear uniforms to school. The telephone poll of 325 adults found that 35% of the adults believe that public high school students should wear uniforms to school. Determine the 95% confidence interval for the true proportion of adults in the county who believe public high school students should wear uniforms.
The 95% confidence interval is % to %.
What is the margin of error?
The margin of error is %. Enter each number as a percent rounded to one decimal. Do not type in the % sign..
I got 29.9 to 40.3 as the interval and 5.2 as the margin of error. Would my answer be correct?
p = .35
n = 325
z = 1.96
E = ('z critical value') * sqrt[p * (1 - p)/n]
E = 1.96 * sqrt [.35* .65/325]
E = 0.0519
p ± E = [.2891, .4019]
The 95% confidence interval is 29.8% to 40.2%.
The margin of error is 5.2.
With a reliability coefficient of 0.81, using classical test theory we would interpret that the:
Answer
a.Amount of error variance to observed variance is 81 percent
b.Amount of true variance to observed variance is 81 percent
c.Instrument has good enough reliability
d.Instrument's validity coefficient would be 0.812
Amount of true variance to observed variance is 81 percent
with the reliability coefficent of 0.81 using the classical theory we would interpret that the
To determine the 95% confidence interval for the true proportion of adults in the county who believe public high school students should wear uniforms, we can use the following formula:
Confidence Interval = Sample Proportion ± Margin of Error
In this case, the sample proportion is 35%, which represents the proportion of adults in the survey who believe that students should wear uniforms.
To calculate the margin of error, we can use the formula:
Margin of Error = Critical Value * Standard Error
The standard error can be calculated using the formula:
Standard Error = sqrt((Sample Proportion * (1 - Sample Proportion)) / Sample Size)
For a 95% confidence level, the critical value is approximately 1.96.
Let's plug in the values:
Sample Proportion = 35%
Sample Size = 325
Critical Value = 1.96
Standard Error = sqrt((0.35 * (1 - 0.35)) / 325) ≈ 0.0238
Margin of Error = 1.96 * 0.0238 ≈ 0.0468 ≈ 4.7% (rounded to one decimal)
Now, let's calculate the confidence interval:
Lower Bound = Sample Proportion - Margin of Error = 35% - 4.7% = 30.3%
Upper Bound = Sample Proportion + Margin of Error = 35% + 4.7% = 39.7%
Therefore, the 95% confidence interval for the true proportion of adults in the county who believe public high school students should wear uniforms is from 30.3% to 39.7%.
The margin of error is 4.7%.
So, your answer of 29.9% to 40.3% as the interval and 5.2% as the margin of error is incorrect. The correct interval is 30.3% to 39.7% with a margin of error of 4.7%.